Differential Equations 1 Question 4
4. If $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$ and $y(0)=1$, then $y \frac{\pi}{2}$ is equal to
(a) $\frac{1}{3}$
(b) $-\frac{2}{3}$
(c) $-\frac{1}{3}$
(d) $\frac{4}{3}$
(2017 Main)
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Answer:
Correct Answer: 4. (a)
Solution:
- We have, $(2+\sin x) \frac{d y}{d x}+(y+1) \cos x=0$
$$ \Rightarrow \quad \frac{d y}{d x}+\frac{\cos x}{2+\sin x} y=\frac{-\cos x}{2+\sin x} $$
which is a linear differential equation.
$\therefore \quad$ IF $=e^{\int \frac{\cos x}{2+\sin x} d x}=e^{\log (2+\sin x)}=2+\sin x$
$\therefore$ Required solution is given by
$$ \begin{aligned} y \cdot(2+\sin x) & =\int \frac{-\cos x}{2+\sin x} \cdot(2+\sin x) d x+C \\ \Rightarrow \quad y(2+\sin x) & =-\sin x+C \end{aligned} $$
Also, $\quad y(0)=1$
$\therefore \quad 1(2+\sin 0)=-\sin 0+C$
$$ \Rightarrow \quad C=2 $$
$\therefore \quad y=\frac{2-\sin x}{2+\sin x} \Rightarrow y \frac{\pi}{2}=\frac{2-\sin \frac{\pi}{2}}{2+\sin \frac{\pi}{2}}=\frac{1}{3}$
